Cartesian Product: \mathbb{R}^3

[quasar987]

Is it true that

[tex]\mathbb{R}\times \mathbb{R}^2 = \mathbb{R}^2 \times \mathbb{R} = \mathbb{R}^3[/tex]

?

 

[Hurkyl]

Yes. (with a but)

 

[BicycleTree]

Wouldn't the elements of [tex]\mathbb{R}\times\mathbb{R}^2[/tex] all be of the form (a, (b, c)) whereas the elements of [tex]\mathbb{R}^2\times\mathbb{R}[/tex] all be of the form ((a, b), c)?

 

[BicycleTree]

(Assuming you meant the 2 to have precedence over the x)

 

[Hurkyl]

That's the essense of the "but".

 

[quasar987]

But how important is this "but"?

 

[Hurkyl]

It's so unimportant, I didn't feel the need to elaborate on it in my response.

In the strictest sense, they aren't equal, but I don't think I've ever really seen anyone distinguish between them.

 

[matt grime]

Strictly speaking they are isomorphic not equal, however the isomoprphism is fairly 'canonical' and in general there are canonical isomorphisms between Ax(BxC) and (AxB)xC, and we will by commonly accepted abuse of notation refer to it as AxBxC.

This is one of the "modern" ways of saying it in the lagauge of category theory.